I'm studyin,g for the first time, the Tensor Product of Vector Spaces. After the answer of a particular question, $[1]$, I think that I grasped the key point of the role of Quotient Vector Space; the answer is:
It's literally immediately from the definition of quotient. If $V/W$ is a quotient space then for $x∈V$ we have $[x]=0$ in $V/W$ $\iff$ $x∈W$.
So $\otimes((v,w_1+w_2)−(v,w_1)−(v,w_2))=0$ in $V⊗W=C(V×W)/Z$ since $(v,w_1+w_2)−(v,w_1)−(v,w_2)∈Z$.
My doubt is: Am I correct on my reasoning written in the following?
Well, you need then some ingredients: Bilinear Maps, and Vector Spaces. You want to construct a general way to think about products of vectors, i.e., you want to construct a single object which contains vectors that lies on different vector spaces. Furthermore, you want to give meaning to components of tensors like $T^{\mu\nu}$, i.e., you want to say formally (construct a set) what is the Tensor object T with components $T^{\mu\nu}$.
In order to do that you need to construct the vector space called THE tensor product $V \otimes W$. This vector space then requires that you construct a bilinear map $\otimes$ which have just "bilinear property and nothing more" $[2]$. Then in order to do that, you must to "use" four vector spaces:
$ V \times W$
$ F(V\times W)$
$ S \subset F(V\times W)$
$ \frac{F(V\times W)}{S}$
With 1. you state the elements $(v,w)$ that will soon be "imaged" into tensors.
With 2. you provide an way to form a linear combination with elements of the set $V \times W$:
$$ u = c^{1}(v_1,w_1) + ... + c^{n}(v_n,w_n) \tag{1}$$
with the caution that $+$ isn't a vector addition, but an way to give meaning of a "single summed entity $u$, with Basis constituted of elements of $V\times W$ " (much like the symbol of complex number written as $a+bi$); $(1)$ forms then the Free Vector Space.
With 3. You choose inteligently the particular vectors of $S \subset F(V\times W)$ to be imaged into something that will provide ("induce") the billinear relationship between elements in $V \otimes W$
With 4., finally, you confirm the necessity of the particular form of the vectors in $S$, because now with the algebraic nature of Quotient Spaces, you can say that:
$$\otimes((v,w_1+w_2)−(v,w_1)−(v,w_2))=0$$ $$\otimes((v_1+v_2,w)−(v_1,w)−(v_2,w))=0$$ $$\otimes(a(v,w)−(av,w))=0$$ $$\otimes(a(v,w)−(v,aw))=0$$
therefore the induced bilinear nature for operation $\otimes$.
In general,then, an element $(v,w)$ will be mapped to $V\otimes W$ by a well defined bilinear map $\otimes$ as the well defined element $\otimes(v,w) \equiv v \otimes w$. By this construction the tensors $v \otimes w$ happens to be equivalence classes of respective element $(v,w)$; by this construction the ideia of a single element that encodes the product of vectors are provided by the element $v \otimes w$ in the Tensor Product:
$$ \frac{F(V\times W)}{S} := V \otimes W $$
$$* * *$$
$[1]$ Tensor Product of Vector Spaces - Quotient Definition
$[2]$ ROMAN.S; Advanced Linear Algebra